andrew_gelman_stats andrew_gelman_stats-2012 andrew_gelman_stats-2012-1333 knowledge-graph by maker-knowledge-mining
Source: html
Introduction: 10. Out of a random sample of 100 Americans, zero report having ever held political office. From this information, give a 95% confidence interval for the proportion of Americans who have ever held political office. Solution to question 9 From yesterday : 9. Out of a population of 100 medical records, 40 are randomly sampled and then audited. 10 out of the 40 audits reveal fraud. From this information, give an estimate, standard error, and 95% confidence interval for the proportion of audits in the population with fraud. Solution: estimate is p.hat=10/40=0.25. Se is sqrt(1-f)*sqrt(p.hat*(1-.hat)/n)=sqrt(1-0.4)*sqrt(0.25*0.75/40)=0.053. 95% interval is [0.25 +/- 2*0.053] = [0.14,0.36].
sentIndex sentText sentNum sentScore
1 Out of a random sample of 100 Americans, zero report having ever held political office. [sent-2, score-0.772]
2 From this information, give a 95% confidence interval for the proportion of Americans who have ever held political office. [sent-3, score-1.338]
3 Solution to question 9 From yesterday : 9. [sent-4, score-0.146]
4 Out of a population of 100 medical records, 40 are randomly sampled and then audited. [sent-5, score-0.496]
5 From this information, give an estimate, standard error, and 95% confidence interval for the proportion of audits in the population with fraud. [sent-7, score-1.444]
wordName wordTfidf (topN-words)
[('sqrt', 0.564), ('audits', 0.365), ('interval', 0.346), ('held', 0.225), ('proportion', 0.216), ('confidence', 0.194), ('solution', 0.183), ('americans', 0.18), ('population', 0.152), ('se', 0.149), ('ever', 0.141), ('sampled', 0.135), ('records', 0.127), ('estimate', 0.126), ('reveal', 0.117), ('randomly', 0.116), ('political', 0.109), ('give', 0.107), ('information', 0.106), ('yesterday', 0.098), ('medical', 0.093), ('zero', 0.081), ('random', 0.077), ('error', 0.071), ('report', 0.071), ('sample', 0.068), ('standard', 0.064), ('question', 0.048)]
simIndex simValue blogId blogTitle
same-blog 1 1.0 1333 andrew gelman stats-2012-05-20-Question 10 of my final exam for Design and Analysis of Sample Surveys
Introduction: 10. Out of a random sample of 100 Americans, zero report having ever held political office. From this information, give a 95% confidence interval for the proportion of Americans who have ever held political office. Solution to question 9 From yesterday : 9. Out of a population of 100 medical records, 40 are randomly sampled and then audited. 10 out of the 40 audits reveal fraud. From this information, give an estimate, standard error, and 95% confidence interval for the proportion of audits in the population with fraud. Solution: estimate is p.hat=10/40=0.25. Se is sqrt(1-f)*sqrt(p.hat*(1-.hat)/n)=sqrt(1-0.4)*sqrt(0.25*0.75/40)=0.053. 95% interval is [0.25 +/- 2*0.053] = [0.14,0.36].
2 0.58781606 1334 andrew gelman stats-2012-05-21-Question 11 of my final exam for Design and Analysis of Sample Surveys
Introduction: 11. Here is the result of fitting a logistic regression to Republican vote in the 1972 NES. Income is on a 1–5 scale. Approximately how much more likely is a person in income category 4 to vote Republican, compared to a person income category 2? Give an approximate estimate, standard error, and 95% interval. Solution to question 10 From yesterday : 10. Out of a random sample of 100 Americans, zero report having ever held political office. From this information, give a 95% confidence interval for the proportion of Americans who have ever held political office. Solution: Use the Agresti-Coull interval based on (y+2)/(n+4). Estimate is p.hat=2/104=0.02, se is sqrt(p.hat*(1-p.hat)/104)=0.013, 95% interval is [0.02 +/- 2*0.013] = [0,0.05].
3 0.39417928 1361 andrew gelman stats-2012-06-02-Question 23 of my final exam for Design and Analysis of Sample Surveys
Introduction: 23. Suppose you are conducting a survey in which people are asked about their health behaviors (how often they wash their hands, how often they go to the doctor, etc.). There is a concern that different interviewers will get different sorts of responses—that is, there may be important interviewer effects. Describe (in two sentences) how you could estimate the interviewer effects within your survey. Can the interviewer effects create problems of reliability of the survey responses? Explain (in one sentence). Can the interviewer effects create problems of validity of the survey responses? Explain (in one sentence). Solution to question 22 From yesterday : 22. A supermarket chain has 100 equally-sized stores. It is desired to estimate the proportion of vegetables that spoil before being sold. Three stores are selected at random and are checked: the percent of spoiled vegetables are 3%, 5%, and 10% in the three stores. Give an estimate and standard error for the percentage of spo
4 0.35361463 1331 andrew gelman stats-2012-05-19-Question 9 of my final exam for Design and Analysis of Sample Surveys
Introduction: 9. Out of a population of 100 medical records, 40 are randomly sampled and then audited. 10 out of the 40 audits reveal fraud. From this information, give an estimate, standard error, and 95% confidence interval for the proportion of audits in the population with fraud. Solution to question 8 From yesterday : 8. Which of the following statements accurately characterize the National Election Studies? (Indicate all that apply.) (a) The NES began in 1960. (b) Since 1980, the NES has mostly relied on telephone interviews. (c) The NES typically has a sample size of about 1000–2000 people. (d) The NES uses a sampling design that ensures they get respondents from all fifty states and D.C. Solution: c. This is a purely factual question, not much to say here.
5 0.29506236 1326 andrew gelman stats-2012-05-17-Question 7 of my final exam for Design and Analysis of Sample Surveys
Introduction: 7. Which of the following statements accurately summarize claims made by Page and Shapiro in The Rational Public and their associated research articles? (Indicate all that apply.) (a) Americans’ attitudes on policy alternatives are highly unstable over time, reflecting a rational response to unstable political conditions. (b) When studying public opinion, question-wording is less important than scholars have traditionally thought. (c) Attitudes about foreign policy change more abruptly than attitudes on domestic issues. (d) The contents of the mass media account for a high proportion of opinion changes on foreign policy. (e) Using the assumption of rationality, Page and Shapiro fit a hedonic regression to estimate the underlying utility function of survey respondents. (f) Page and Shapiro use the term “rational” ironically; their fundamental claim is that Americans are easily distracted and that rational-public models are seriously flawed. Solution to question 6 From
6 0.27510232 480 andrew gelman stats-2010-12-21-Instead of “confidence interval,” let’s say “uncertainty interval”
8 0.2471444 1349 andrew gelman stats-2012-05-28-Question 18 of my final exam for Design and Analysis of Sample Surveys
10 0.22191001 1352 andrew gelman stats-2012-05-29-Question 19 of my final exam for Design and Analysis of Sample Surveys
11 0.21284543 1672 andrew gelman stats-2013-01-14-How do you think about the values in a confidence interval?
12 0.14306532 1337 andrew gelman stats-2012-05-22-Question 12 of my final exam for Design and Analysis of Sample Surveys
13 0.12840316 1320 andrew gelman stats-2012-05-14-Question 4 of my final exam for Design and Analysis of Sample Surveys
14 0.12593837 2201 andrew gelman stats-2014-02-06-Bootstrap averaging: Examples where it works and where it doesn’t work
15 0.12562737 1341 andrew gelman stats-2012-05-24-Question 14 of my final exam for Design and Analysis of Sample Surveys
16 0.11901257 1358 andrew gelman stats-2012-06-01-Question 22 of my final exam for Design and Analysis of Sample Surveys
17 0.11272058 1348 andrew gelman stats-2012-05-27-Question 17 of my final exam for Design and Analysis of Sample Surveys
18 0.10955369 1315 andrew gelman stats-2012-05-12-Question 2 of my final exam for Design and Analysis of Sample Surveys
19 0.10935068 1941 andrew gelman stats-2013-07-16-Priors
topicId topicWeight
[(0, 0.083), (1, 0.043), (2, 0.173), (3, -0.094), (4, 0.032), (5, 0.033), (6, 0.004), (7, 0.074), (8, -0.017), (9, -0.198), (10, 0.016), (11, -0.086), (12, -0.012), (13, 0.124), (14, -0.075), (15, -0.09), (16, -0.103), (17, -0.048), (18, 0.06), (19, -0.086), (20, 0.135), (21, -0.062), (22, 0.115), (23, -0.008), (24, 0.169), (25, -0.099), (26, -0.008), (27, -0.095), (28, -0.088), (29, -0.013), (30, -0.081), (31, -0.071), (32, -0.043), (33, -0.099), (34, 0.092), (35, 0.013), (36, -0.073), (37, 0.107), (38, 0.097), (39, -0.034), (40, -0.026), (41, -0.045), (42, 0.055), (43, 0.056), (44, -0.018), (45, 0.033), (46, -0.032), (47, -0.042), (48, -0.008), (49, -0.028)]
simIndex simValue blogId blogTitle
same-blog 1 0.99441808 1333 andrew gelman stats-2012-05-20-Question 10 of my final exam for Design and Analysis of Sample Surveys
Introduction: 10. Out of a random sample of 100 Americans, zero report having ever held political office. From this information, give a 95% confidence interval for the proportion of Americans who have ever held political office. Solution to question 9 From yesterday : 9. Out of a population of 100 medical records, 40 are randomly sampled and then audited. 10 out of the 40 audits reveal fraud. From this information, give an estimate, standard error, and 95% confidence interval for the proportion of audits in the population with fraud. Solution: estimate is p.hat=10/40=0.25. Se is sqrt(1-f)*sqrt(p.hat*(1-.hat)/n)=sqrt(1-0.4)*sqrt(0.25*0.75/40)=0.053. 95% interval is [0.25 +/- 2*0.053] = [0.14,0.36].
2 0.89479387 1334 andrew gelman stats-2012-05-21-Question 11 of my final exam for Design and Analysis of Sample Surveys
Introduction: 11. Here is the result of fitting a logistic regression to Republican vote in the 1972 NES. Income is on a 1–5 scale. Approximately how much more likely is a person in income category 4 to vote Republican, compared to a person income category 2? Give an approximate estimate, standard error, and 95% interval. Solution to question 10 From yesterday : 10. Out of a random sample of 100 Americans, zero report having ever held political office. From this information, give a 95% confidence interval for the proportion of Americans who have ever held political office. Solution: Use the Agresti-Coull interval based on (y+2)/(n+4). Estimate is p.hat=2/104=0.02, se is sqrt(p.hat*(1-p.hat)/104)=0.013, 95% interval is [0.02 +/- 2*0.013] = [0,0.05].
3 0.87939817 1331 andrew gelman stats-2012-05-19-Question 9 of my final exam for Design and Analysis of Sample Surveys
Introduction: 9. Out of a population of 100 medical records, 40 are randomly sampled and then audited. 10 out of the 40 audits reveal fraud. From this information, give an estimate, standard error, and 95% confidence interval for the proportion of audits in the population with fraud. Solution to question 8 From yesterday : 8. Which of the following statements accurately characterize the National Election Studies? (Indicate all that apply.) (a) The NES began in 1960. (b) Since 1980, the NES has mostly relied on telephone interviews. (c) The NES typically has a sample size of about 1000–2000 people. (d) The NES uses a sampling design that ensures they get respondents from all fifty states and D.C. Solution: c. This is a purely factual question, not much to say here.
4 0.72735107 480 andrew gelman stats-2010-12-21-Instead of “confidence interval,” let’s say “uncertainty interval”
Introduction: I’ve become increasingly uncomfortable with the term “confidence interval,” for several reasons: - The well-known difficulties in interpretation (officially the confidence statement can be interpreted only on average, but people typically implicitly give the Bayesian interpretation to each case), - The ambiguity between confidence intervals and predictive intervals. (See the footnote in BDA where we discuss the difference between “inference” and “prediction” in the classical framework.) - The awkwardness of explaining that confidence intervals are big in noisy situations where you have less confidence, and confidence intervals are small when you have more confidence. So here’s my proposal. Let’s use the term “uncertainty interval” instead. The uncertainty interval tells you how much uncertainty you have. That works pretty well, I think. P.S. As of this writing, “confidence interval” outGoogles “uncertainty interval” by the huge margin of 9.5 million to 54000. So we
5 0.70247597 1672 andrew gelman stats-2013-01-14-How do you think about the values in a confidence interval?
Introduction: Philip Jones writes: As an interested reader of your blog, I wondered if you might consider a blog entry sometime on the following question I posed on CrossValidated (StackExchange). I originally posed the question based on my uncertainty about 95% CIs: “Are all values within the 95% CI equally likely (probable), or are the values at the “tails” of the 95% CI less likely than those in the middle of the CI closer to the point estimate?” I posed this question based on discordant information I found at a couple of different web sources (I posted these sources in the body of the question). I received some interesting replies, and the replies were not unanimous, in fact there is some serious disagreement there! After seeing this disagreement, I naturally thought of you, and whether you might be able to clear this up. Please note I am not referring to credible intervals, but rather to the common medical journal reporting standard of confidence intervals. My response: First
6 0.67459929 1361 andrew gelman stats-2012-06-02-Question 23 of my final exam for Design and Analysis of Sample Surveys
7 0.64534074 1348 andrew gelman stats-2012-05-27-Question 17 of my final exam for Design and Analysis of Sample Surveys
8 0.62389392 1337 andrew gelman stats-2012-05-22-Question 12 of my final exam for Design and Analysis of Sample Surveys
9 0.61752391 1358 andrew gelman stats-2012-06-01-Question 22 of my final exam for Design and Analysis of Sample Surveys
10 0.61409944 1326 andrew gelman stats-2012-05-17-Question 7 of my final exam for Design and Analysis of Sample Surveys
11 0.60064012 1349 andrew gelman stats-2012-05-28-Question 18 of my final exam for Design and Analysis of Sample Surveys
13 0.57168531 2201 andrew gelman stats-2014-02-06-Bootstrap averaging: Examples where it works and where it doesn’t work
14 0.56966043 1320 andrew gelman stats-2012-05-14-Question 4 of my final exam for Design and Analysis of Sample Surveys
15 0.55498123 1345 andrew gelman stats-2012-05-26-Question 16 of my final exam for Design and Analysis of Sample Surveys
17 0.53589696 1328 andrew gelman stats-2012-05-18-Question 8 of my final exam for Design and Analysis of Sample Surveys
19 0.50410515 1340 andrew gelman stats-2012-05-23-Question 13 of my final exam for Design and Analysis of Sample Surveys
20 0.50362885 1362 andrew gelman stats-2012-06-03-Question 24 of my final exam for Design and Analysis of Sample Surveys
topicId topicWeight
[(16, 0.057), (24, 0.035), (27, 0.026), (65, 0.33), (99, 0.374)]
simIndex simValue blogId blogTitle
1 0.96906042 457 andrew gelman stats-2010-12-07-Whassup with phantom-limb treatment?
Introduction: OK, here’s something that is completely baffling me. I read this article by John Colapinto on the neuroscientist V. S. Ramachandran, who’s famous for his innovative treatment for “phantom limb” pain: His first subject was a young man who a decade earlier had crashed his motorcycle and torn from his spinal column the nerves supplying the left arm. After keeping the useless arm in a sling for a year, the man had the arm amputated above the elbow. Ever since, he had felt unremitting cramping in the phantom limb, as though it were immobilized in an awkward position. . . . Ramachandram positioned a twenty-inch-by-twenty-inch drugstore mirror . . . and told him to place his intact right arm on one side of the mirror and his stump on the other. He told the man to arrange the mirror so that the reflection created the illusion that his intact arm was the continuation of the amputated one. The Ramachandran asked the man to move his right and left arms . . . “Oh, my God!” the man began
same-blog 2 0.96153259 1333 andrew gelman stats-2012-05-20-Question 10 of my final exam for Design and Analysis of Sample Surveys
Introduction: 10. Out of a random sample of 100 Americans, zero report having ever held political office. From this information, give a 95% confidence interval for the proportion of Americans who have ever held political office. Solution to question 9 From yesterday : 9. Out of a population of 100 medical records, 40 are randomly sampled and then audited. 10 out of the 40 audits reveal fraud. From this information, give an estimate, standard error, and 95% confidence interval for the proportion of audits in the population with fraud. Solution: estimate is p.hat=10/40=0.25. Se is sqrt(1-f)*sqrt(p.hat*(1-.hat)/n)=sqrt(1-0.4)*sqrt(0.25*0.75/40)=0.053. 95% interval is [0.25 +/- 2*0.053] = [0.14,0.36].
3 0.93785906 990 andrew gelman stats-2011-11-04-At the politics blogs . . .
Introduction: There is no increased inequality, unless you look at the top 1% The revolving door of U.S. politics The redistricting song
4 0.92306376 1993 andrew gelman stats-2013-08-22-Improvements to Kindle Version of BDA3
Introduction: I let Andrew know about the comments about the defective Kindle version of BDA2 and he wrote to his editor at Chapman and Hall, Rob Calver, who wrote back with this info: I can guarantee that the Kindle version of the third edition will be a substantial improvement. We publish all of our mathematics and statistics books through Kindle now as Print Replica. This means that we send the printer pdf to Amazon and they convert into their Print Replica format, which is essentially just a pdf viewer. We have not experienced very many issues at all with this setup. Unfortunately, there was a period before Amazon launched Print Replica when they converted math/stat books into their Kindle format, and converted them very badly in some cases. Equations were held as images, making them very difficult to read. It appears this was the case with Andrew’s second edition, judging by some of the comments. The third edition [of BDA] will be available through Kindle with a short delay (for Amazo
5 0.91393924 396 andrew gelman stats-2010-11-05-Journalism in the age of data
Introduction: Journalism in the age of data is a video report including interviews with many visualization people. It’s also a great example of how citations, and further information appear alongside with the video – showing us the future of video content online.
6 0.8848269 2062 andrew gelman stats-2013-10-15-Last word on Mister P (for now)
7 0.87516564 1021 andrew gelman stats-2011-11-21-Don’t judge a book by its title
8 0.86216372 1845 andrew gelman stats-2013-05-07-Is Felix Salmon wrong on free TV?
9 0.85310626 1197 andrew gelman stats-2012-03-04-“All Models are Right, Most are Useless”
10 0.85145223 100 andrew gelman stats-2010-06-19-Unsurprisingly, people are more worried about the economy and jobs than about deficits
11 0.84771109 1426 andrew gelman stats-2012-07-23-Special effects
12 0.84190166 1361 andrew gelman stats-2012-06-02-Question 23 of my final exam for Design and Analysis of Sample Surveys
13 0.83992338 2056 andrew gelman stats-2013-10-09-Mister P: What’s its secret sauce?
14 0.82686353 1334 andrew gelman stats-2012-05-21-Question 11 of my final exam for Design and Analysis of Sample Surveys
15 0.82571596 2292 andrew gelman stats-2014-04-15-When you believe in things that you don’t understand
16 0.82415962 2061 andrew gelman stats-2013-10-14-More on Mister P and how it does what it does
17 0.82300323 1365 andrew gelman stats-2012-06-04-Question 25 of my final exam for Design and Analysis of Sample Surveys
18 0.81766886 1475 andrew gelman stats-2012-08-30-A Stan is Born
19 0.81698072 671 andrew gelman stats-2011-04-20-One more time-use graph
20 0.81694251 463 andrew gelman stats-2010-12-11-Compare p-values from privately funded medical trials to those in publicly funded research?